NEET Test Series from KOTA - 10 Papers In MS WORD
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Linear Inequalities and Linear Programming
88543
The maximum value \(z=6 x+8 y\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) is
1 42
2 72
3 24
4 96
Explanation:
(B) Given, \(\mathrm{z}=6 \mathrm{x}+8 \mathrm{y}\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) | In \lt br> equati \lt br> on | Equati \lt br> on | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Regi \lt br> on | | :---: | :---: | :---: | :---: | :---: | :---: | | $x-y \geq 0$ | $x-y=0$ | 1 | 1 | $(1,1)$ | - | | | | 4 | 4 | $(4,4)$ | | | $x+3 y \leq 1$ | $x+3 y=1$ | 12 | 0 | $(12,0)$ | Origin \lt br> side | | | | 0 | 4 | $(0,4)$ | | Diagram The shaded region is the feasible region. The corner points are \(\mathrm{O}(0,0), \mathrm{A}(12,0), \mathrm{B}(3,3)\) | Points | $\mathbf{z}=\mathbf{6 x}+\mathbf{8 y}$ | | :--- | :--- | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(12,0)$ | 72 | | $\mathrm{B}(3,3)$ | 42 | \(\therefore \mathrm{z}\) has maximum value 72 at \(\mathrm{A}(12,0)\).
MHT CET-2019
Linear Inequalities and Linear Programming
88544
If \(z=a x+b y ; a, b>0\) subject to \(x \leq 2, y \leq 2, x\) \(+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq \mathbf{0}\) has minimum value at \((2,1)\) only, then
88545
The maximum value of \(z=75 x+50 y\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) is
1 575
2 600
3 580
4 400
Explanation:
(A) : Given, \(\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) | In equation | Equation | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Region | | :---: | :---: | :---: | :---: | :---: | :---: | | $8 \mathrm{x}+5 \mathrm{y} \leq 60$ | $8 \mathrm{x}+5 \mathrm{y}=60$ | 7.5 | 0 | $(7.5,0)$ | Origin | | | | 0 | 12 | $(0,12)$ | side | | $4 \mathrm{x}+5 \mathrm{y} \leq 40$ | $4 \mathrm{x}+5 \mathrm{y}=40$ | 10 | 0 | $(10,0)$ | Origin | | | | 0 | 8 | $(0,8)$ | side | The shaded part is the feasible region. | Points | $\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}$ | | :---: | :---: | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(7.5,0)$ | 562.5 | | $\mathrm{B}(5,4)$ | 575 | | $\mathrm{C}(0,8)$ | 400 | | $\mathrm{z}$ has minimum value at $\mathrm{B}(5,4)$ is 575 | |
MHT CET-2019
Linear Inequalities and Linear Programming
88546
The minimum value of \(z=5 x+4 y\) subject to \(\mathbf{y} \leq \mathbf{2 x}, \mathrm{x} \leq \mathbf{2} \mathbf{y}, \mathbf{x}+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq 0\) is
1 12
2 14
3 13
4 10
Explanation:
(C) : Given, \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) subject is \(y \leq 2 x, \quad x \leq 2 y, \quad x+y \geq 3, x \geq 0, \quad y \geq 0\) Let us calculate point \(B\). solve \(y=2 x\) and \(x+y=3\) \(2 \mathrm{x}-\mathrm{y}=0\) \(\mathrm{x}+\mathrm{y}=3\) \(3 \mathrm{x}=3\) \(\mathrm{x}=1 \quad\) So, the point \(\mathrm{B}\) is \((1,2)\) \(\mathrm{y}=2\) Calculate point \(\mathrm{C}:-\) Solve \(2 \mathrm{y}-\mathrm{x}=0\) and \(\mathrm{x}+\mathrm{y}=3\) We get \(\mathrm{y}=1\) \(x=2\) \(\therefore\) The point is \((2,1)\) Point D is \((3,0)\) \(\therefore\) value of \(\mathrm{z}\) corresponding to point \(\mathrm{A}\) is \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) \(=0+12=12\) Point \(\mathrm{B}, \mathrm{z}=5 \times 1+4 \times 2=13\) Point \(\mathrm{c} \mathrm{z}=2 \times 5+1 \times 4=14\) Points \(\mathrm{B}\) and \(\mathrm{C}\) are considered \(\therefore\) The minimum value is 13 .
88543
The maximum value \(z=6 x+8 y\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) is
1 42
2 72
3 24
4 96
Explanation:
(B) Given, \(\mathrm{z}=6 \mathrm{x}+8 \mathrm{y}\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) | In \lt br> equati \lt br> on | Equati \lt br> on | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Regi \lt br> on | | :---: | :---: | :---: | :---: | :---: | :---: | | $x-y \geq 0$ | $x-y=0$ | 1 | 1 | $(1,1)$ | - | | | | 4 | 4 | $(4,4)$ | | | $x+3 y \leq 1$ | $x+3 y=1$ | 12 | 0 | $(12,0)$ | Origin \lt br> side | | | | 0 | 4 | $(0,4)$ | | Diagram The shaded region is the feasible region. The corner points are \(\mathrm{O}(0,0), \mathrm{A}(12,0), \mathrm{B}(3,3)\) | Points | $\mathbf{z}=\mathbf{6 x}+\mathbf{8 y}$ | | :--- | :--- | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(12,0)$ | 72 | | $\mathrm{B}(3,3)$ | 42 | \(\therefore \mathrm{z}\) has maximum value 72 at \(\mathrm{A}(12,0)\).
MHT CET-2019
Linear Inequalities and Linear Programming
88544
If \(z=a x+b y ; a, b>0\) subject to \(x \leq 2, y \leq 2, x\) \(+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq \mathbf{0}\) has minimum value at \((2,1)\) only, then
88545
The maximum value of \(z=75 x+50 y\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) is
1 575
2 600
3 580
4 400
Explanation:
(A) : Given, \(\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) | In equation | Equation | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Region | | :---: | :---: | :---: | :---: | :---: | :---: | | $8 \mathrm{x}+5 \mathrm{y} \leq 60$ | $8 \mathrm{x}+5 \mathrm{y}=60$ | 7.5 | 0 | $(7.5,0)$ | Origin | | | | 0 | 12 | $(0,12)$ | side | | $4 \mathrm{x}+5 \mathrm{y} \leq 40$ | $4 \mathrm{x}+5 \mathrm{y}=40$ | 10 | 0 | $(10,0)$ | Origin | | | | 0 | 8 | $(0,8)$ | side | The shaded part is the feasible region. | Points | $\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}$ | | :---: | :---: | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(7.5,0)$ | 562.5 | | $\mathrm{B}(5,4)$ | 575 | | $\mathrm{C}(0,8)$ | 400 | | $\mathrm{z}$ has minimum value at $\mathrm{B}(5,4)$ is 575 | |
MHT CET-2019
Linear Inequalities and Linear Programming
88546
The minimum value of \(z=5 x+4 y\) subject to \(\mathbf{y} \leq \mathbf{2 x}, \mathrm{x} \leq \mathbf{2} \mathbf{y}, \mathbf{x}+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq 0\) is
1 12
2 14
3 13
4 10
Explanation:
(C) : Given, \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) subject is \(y \leq 2 x, \quad x \leq 2 y, \quad x+y \geq 3, x \geq 0, \quad y \geq 0\) Let us calculate point \(B\). solve \(y=2 x\) and \(x+y=3\) \(2 \mathrm{x}-\mathrm{y}=0\) \(\mathrm{x}+\mathrm{y}=3\) \(3 \mathrm{x}=3\) \(\mathrm{x}=1 \quad\) So, the point \(\mathrm{B}\) is \((1,2)\) \(\mathrm{y}=2\) Calculate point \(\mathrm{C}:-\) Solve \(2 \mathrm{y}-\mathrm{x}=0\) and \(\mathrm{x}+\mathrm{y}=3\) We get \(\mathrm{y}=1\) \(x=2\) \(\therefore\) The point is \((2,1)\) Point D is \((3,0)\) \(\therefore\) value of \(\mathrm{z}\) corresponding to point \(\mathrm{A}\) is \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) \(=0+12=12\) Point \(\mathrm{B}, \mathrm{z}=5 \times 1+4 \times 2=13\) Point \(\mathrm{c} \mathrm{z}=2 \times 5+1 \times 4=14\) Points \(\mathrm{B}\) and \(\mathrm{C}\) are considered \(\therefore\) The minimum value is 13 .
88543
The maximum value \(z=6 x+8 y\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) is
1 42
2 72
3 24
4 96
Explanation:
(B) Given, \(\mathrm{z}=6 \mathrm{x}+8 \mathrm{y}\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) | In \lt br> equati \lt br> on | Equati \lt br> on | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Regi \lt br> on | | :---: | :---: | :---: | :---: | :---: | :---: | | $x-y \geq 0$ | $x-y=0$ | 1 | 1 | $(1,1)$ | - | | | | 4 | 4 | $(4,4)$ | | | $x+3 y \leq 1$ | $x+3 y=1$ | 12 | 0 | $(12,0)$ | Origin \lt br> side | | | | 0 | 4 | $(0,4)$ | | Diagram The shaded region is the feasible region. The corner points are \(\mathrm{O}(0,0), \mathrm{A}(12,0), \mathrm{B}(3,3)\) | Points | $\mathbf{z}=\mathbf{6 x}+\mathbf{8 y}$ | | :--- | :--- | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(12,0)$ | 72 | | $\mathrm{B}(3,3)$ | 42 | \(\therefore \mathrm{z}\) has maximum value 72 at \(\mathrm{A}(12,0)\).
MHT CET-2019
Linear Inequalities and Linear Programming
88544
If \(z=a x+b y ; a, b>0\) subject to \(x \leq 2, y \leq 2, x\) \(+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq \mathbf{0}\) has minimum value at \((2,1)\) only, then
88545
The maximum value of \(z=75 x+50 y\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) is
1 575
2 600
3 580
4 400
Explanation:
(A) : Given, \(\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) | In equation | Equation | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Region | | :---: | :---: | :---: | :---: | :---: | :---: | | $8 \mathrm{x}+5 \mathrm{y} \leq 60$ | $8 \mathrm{x}+5 \mathrm{y}=60$ | 7.5 | 0 | $(7.5,0)$ | Origin | | | | 0 | 12 | $(0,12)$ | side | | $4 \mathrm{x}+5 \mathrm{y} \leq 40$ | $4 \mathrm{x}+5 \mathrm{y}=40$ | 10 | 0 | $(10,0)$ | Origin | | | | 0 | 8 | $(0,8)$ | side | The shaded part is the feasible region. | Points | $\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}$ | | :---: | :---: | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(7.5,0)$ | 562.5 | | $\mathrm{B}(5,4)$ | 575 | | $\mathrm{C}(0,8)$ | 400 | | $\mathrm{z}$ has minimum value at $\mathrm{B}(5,4)$ is 575 | |
MHT CET-2019
Linear Inequalities and Linear Programming
88546
The minimum value of \(z=5 x+4 y\) subject to \(\mathbf{y} \leq \mathbf{2 x}, \mathrm{x} \leq \mathbf{2} \mathbf{y}, \mathbf{x}+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq 0\) is
1 12
2 14
3 13
4 10
Explanation:
(C) : Given, \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) subject is \(y \leq 2 x, \quad x \leq 2 y, \quad x+y \geq 3, x \geq 0, \quad y \geq 0\) Let us calculate point \(B\). solve \(y=2 x\) and \(x+y=3\) \(2 \mathrm{x}-\mathrm{y}=0\) \(\mathrm{x}+\mathrm{y}=3\) \(3 \mathrm{x}=3\) \(\mathrm{x}=1 \quad\) So, the point \(\mathrm{B}\) is \((1,2)\) \(\mathrm{y}=2\) Calculate point \(\mathrm{C}:-\) Solve \(2 \mathrm{y}-\mathrm{x}=0\) and \(\mathrm{x}+\mathrm{y}=3\) We get \(\mathrm{y}=1\) \(x=2\) \(\therefore\) The point is \((2,1)\) Point D is \((3,0)\) \(\therefore\) value of \(\mathrm{z}\) corresponding to point \(\mathrm{A}\) is \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) \(=0+12=12\) Point \(\mathrm{B}, \mathrm{z}=5 \times 1+4 \times 2=13\) Point \(\mathrm{c} \mathrm{z}=2 \times 5+1 \times 4=14\) Points \(\mathrm{B}\) and \(\mathrm{C}\) are considered \(\therefore\) The minimum value is 13 .
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Linear Inequalities and Linear Programming
88543
The maximum value \(z=6 x+8 y\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) is
1 42
2 72
3 24
4 96
Explanation:
(B) Given, \(\mathrm{z}=6 \mathrm{x}+8 \mathrm{y}\) subject to \(x-y \geq 0, x+3 y \leq 12, x \geq 0, y \geq 0\) | In \lt br> equati \lt br> on | Equati \lt br> on | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Regi \lt br> on | | :---: | :---: | :---: | :---: | :---: | :---: | | $x-y \geq 0$ | $x-y=0$ | 1 | 1 | $(1,1)$ | - | | | | 4 | 4 | $(4,4)$ | | | $x+3 y \leq 1$ | $x+3 y=1$ | 12 | 0 | $(12,0)$ | Origin \lt br> side | | | | 0 | 4 | $(0,4)$ | | Diagram The shaded region is the feasible region. The corner points are \(\mathrm{O}(0,0), \mathrm{A}(12,0), \mathrm{B}(3,3)\) | Points | $\mathbf{z}=\mathbf{6 x}+\mathbf{8 y}$ | | :--- | :--- | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(12,0)$ | 72 | | $\mathrm{B}(3,3)$ | 42 | \(\therefore \mathrm{z}\) has maximum value 72 at \(\mathrm{A}(12,0)\).
MHT CET-2019
Linear Inequalities and Linear Programming
88544
If \(z=a x+b y ; a, b>0\) subject to \(x \leq 2, y \leq 2, x\) \(+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq \mathbf{0}\) has minimum value at \((2,1)\) only, then
88545
The maximum value of \(z=75 x+50 y\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) is
1 575
2 600
3 580
4 400
Explanation:
(A) : Given, \(\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}\), subject to \(8 x+5 y \leq 60,4 x+5 y \leq 40, x \geq 0, y \geq 0\) | In equation | Equation | $\mathbf{X}$ | $\mathbf{y}$ | $(\mathbf{x}, \mathbf{y})$ | Region | | :---: | :---: | :---: | :---: | :---: | :---: | | $8 \mathrm{x}+5 \mathrm{y} \leq 60$ | $8 \mathrm{x}+5 \mathrm{y}=60$ | 7.5 | 0 | $(7.5,0)$ | Origin | | | | 0 | 12 | $(0,12)$ | side | | $4 \mathrm{x}+5 \mathrm{y} \leq 40$ | $4 \mathrm{x}+5 \mathrm{y}=40$ | 10 | 0 | $(10,0)$ | Origin | | | | 0 | 8 | $(0,8)$ | side | The shaded part is the feasible region. | Points | $\mathrm{z}=75 \mathrm{x}+50 \mathrm{y}$ | | :---: | :---: | | $\mathrm{O}(0,0)$ | 0 | | $\mathrm{A}(7.5,0)$ | 562.5 | | $\mathrm{B}(5,4)$ | 575 | | $\mathrm{C}(0,8)$ | 400 | | $\mathrm{z}$ has minimum value at $\mathrm{B}(5,4)$ is 575 | |
MHT CET-2019
Linear Inequalities and Linear Programming
88546
The minimum value of \(z=5 x+4 y\) subject to \(\mathbf{y} \leq \mathbf{2 x}, \mathrm{x} \leq \mathbf{2} \mathbf{y}, \mathbf{x}+\mathbf{y} \geq \mathbf{3}, \mathbf{x} \geq \mathbf{0}, \mathrm{y} \geq 0\) is
1 12
2 14
3 13
4 10
Explanation:
(C) : Given, \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) subject is \(y \leq 2 x, \quad x \leq 2 y, \quad x+y \geq 3, x \geq 0, \quad y \geq 0\) Let us calculate point \(B\). solve \(y=2 x\) and \(x+y=3\) \(2 \mathrm{x}-\mathrm{y}=0\) \(\mathrm{x}+\mathrm{y}=3\) \(3 \mathrm{x}=3\) \(\mathrm{x}=1 \quad\) So, the point \(\mathrm{B}\) is \((1,2)\) \(\mathrm{y}=2\) Calculate point \(\mathrm{C}:-\) Solve \(2 \mathrm{y}-\mathrm{x}=0\) and \(\mathrm{x}+\mathrm{y}=3\) We get \(\mathrm{y}=1\) \(x=2\) \(\therefore\) The point is \((2,1)\) Point D is \((3,0)\) \(\therefore\) value of \(\mathrm{z}\) corresponding to point \(\mathrm{A}\) is \(\mathrm{z}=5 \mathrm{x}+4 \mathrm{y}\) \(=0+12=12\) Point \(\mathrm{B}, \mathrm{z}=5 \times 1+4 \times 2=13\) Point \(\mathrm{c} \mathrm{z}=2 \times 5+1 \times 4=14\) Points \(\mathrm{B}\) and \(\mathrm{C}\) are considered \(\therefore\) The minimum value is 13 .
